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{SECT 0 {PARA 259 "" 0 "" {TEXT -1 31 "APPLICATIONS OF DERIVATIVES - I
" }}{PARA 4 "" 0 "" {TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT -1 18 "Calcu
lus I Project" }}{PARA 0 "" 0 "" {TEXT -1 38 "                        \+
              " }{HYPERLNK 17 "" 1 "" "" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 1 " " }{TEXT 256 12 "Objectives:
 " }}{PARA 0 "" 0 "" {TEXT 261 1 " " }{TEXT 263 0 "" }{TEXT 264 0 "" }
{TEXT -1 1 " " }{TEXT 265 80 "To study average rate of change and inst
antaneous rate of change algebraically, " }{TEXT 262 1 " " }{TEXT 266 
1 " " }}{PARA 0 "" 0 "" {TEXT -1 28 "numerically and graphically." }}
{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 29 "Use a co
mputer algebra system" }{TEXT 267 1 " " }{TEXT -1 42 "to graph, simpli
fy and evaluate functions." }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 1 " " 
}{TEXT 268 15 "Solved Example:" }}{PARA 0 "" 0 "" {TEXT -1 81 "  A tan
k is filled with water by opening a valve on an inlet pipe.  The volum
e V " }}{PARA 0 "" 0 "" {TEXT -1 81 "in liters of water in the tank t \+
min, after the valve is opened, is given by the " }}{PARA 0 "" 0 "" 
{TEXT -1 10 "formula V(" }{TEXT 280 1 "t" }{TEXT -1 5 ") = 5" }
{XPPEDIT 18 0 "t^2;" "6#*$%\"tG\"\"#" }{TEXT -1 4 " + 4" }{TEXT 281 1 
"t" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" {TEXT -1 71 "        a)  Plot \+
the graph of the average rate of change in the volume " }}{PARA 0 "" 
0 "" {TEXT -1 18 "             from " }{TEXT 282 1 "t" }{TEXT -1 12 " \+
= 2 to 2 + " }{TEXT 283 1 "h" }{TEXT -1 44 ".  Use it to estimate the \+
rate of change at " }{TEXT 284 1 "t" }{TEXT -1 4 " =2." }}{PARA 0 "" 
0 "" {TEXT -1 66 "        b)  What is the average rate of change in th
e volume from " }}{PARA 0 "" 0 "" {TEXT -1 14 "              " }{TEXT 
285 1 "t" }{TEXT -1 11 " = 1.99 to " }{TEXT 286 1 "t" }{TEXT -1 10 " =
 2.00,  " }{TEXT 287 1 "t" }{TEXT -1 11 " = 2.00 to " }{TEXT 288 1 "t
" }{TEXT -1 10 " = 2.01,  " }{TEXT 289 1 "t" }{TEXT -1 11 " = 1.99 to \+
" }{TEXT 290 1 "t" }{TEXT -1 8 " = 2.01 " }}{PARA 0 "" 0 "" {TEXT -1 
67 "              minutes.  Use this to estimate the rate of change at
 " }{TEXT 291 1 "t" }{TEXT -1 5 " = 2." }}{PARA 0 "" 0 "" {TEXT -1 55 
"        c)  Find the instantaneous rate of increase at " }{TEXT 292 
1 "t" }{TEXT -1 27 " = 2 by taking appropriate " }}{PARA 0 "" 0 "" 
{TEXT -1 20 "             limit. " }}{PARA 0 "" 0 "" {TEXT -1 45 "    \+
    d)  Evaluate the rate of increase at " }{TEXT 293 1 "t" }{TEXT -1 
38 " = 2 by printing a table of values of " }}{PARA 0 "" 0 "" {TEXT 
-1 13 "             " }{XPPEDIT 18 0 "(V(2+h)-V(2))/h;" "6#*&,&-%\"VG6
#,&\"\"#\"\"\"%\"hGF*F*-F&6#F)!\"\"F*F+F." }{TEXT -1 24 "  for various
 values of " }{TEXT 294 1 "h" }{TEXT -1 23 " close to 0, omitting  " }
{TEXT 295 1 "h" }{TEXT -1 7 " = 0.  " }}{PARA 0 "" 0 "" {TEXT -1 18 " \+
            (Use " }{TEXT 296 1 "h" }{TEXT -1 72 " = -.05, -.04, -.03,
 -.02, -.01, .01, .02, .03, .04, .05).              " }}{PARA 0 "" 0 "
" {TEXT -1 76 "        e)  Compare your answer with the answers in  a)
 ,  b),  c) and  d). " }}{PARA 0 "" 0 "" {TEXT -1 4 "    " }}}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 11 " Solution
: " }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 274 2 "a)" }{TEXT 
-1 1 " " }{TEXT 273 30 "Define function and plot graph" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "V := t -> 5*t^2 + 4*t;  \n" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"VGf*6#%\"tG6\"6$%)operatorG%&arrow
GF(,&*&\"\"&\"\"\")9$\"\"#F/F/*&\"\"%F/F1F/F/F(F(F(" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 13 "with(plots);\n" }}{PARA 12 "" 1 "" 
{XPPMATH 20 "6#7Z%(animateG%*animate3dG%-animatecurveG%&arrowG%-change
coordsG%,complexplotG%.complexplot3dG%*conformalG%,conformal3dG%,conto
urplotG%.contourplot3dG%*coordplotG%,coordplot3dG%-cylinderplotG%,dens
ityplotG%(displayG%*display3dG%*fieldplotG%,fieldplot3dG%)gradplotG%+g
radplot3dG%,graphplot3dG%-implicitplotG%/implicitplot3dG%(inequalG%,in
teractiveG%-listcontplotG%/listcontplot3dG%0listdensityplotG%)listplot
G%+listplot3dG%+loglogplotG%(logplotG%+matrixplotG%(odeplotG%'paretoG%
,plotcompareG%*pointplotG%,pointplot3dG%*polarplotG%,polygonplotG%.pol
ygonplot3dG%4polyhedra_supportedG%.polyhedraplotG%'replotG%*rootlocusG
%,semilogplotG%+setoptionsG%-setoptions3dG%+spacecurveG%1sparsematrixp
lotG%+sphereplotG%)surfdataG%)textplotG%+textplot3dG%)tubeplotG" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 109 "plot((V(2 + h) - V(2))/h, h
 = -1..1, title = `Graph of (V(2 + h) - V(2))/h over the interval h = \+
-1 to 1` );\n" }}{PARA 13 "" 1 "" {GLPLOT2D 409 193 193 {PLOTDATA 2 "6
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*\\7k.6aF0$\"3)3+D1#=bqEF37$$\"3#emmmT9C#eF0$\"3uLL$3s?6p#F37$$\"33***
*\\i!*3`iF0$\"38+]7`Wl7FF37$$\"3%QLLL$*zym'F0$\"3_lmm'*RRLFF37$$\"3wKL
L3N1#4(F0$\"3?lmTvJgaFF37$$\"3Nmm;HYt7vF0$\"3#HLe9tOcx#F37$$\"3Y******
*p(G**yF0$\"35+++&Qk\\z#F37$$\"3]mmmT6KU$)F0$\"3ILL3dg6<GF37$$\"3fKLLL
bdQ()F0$\"3almmw(Gp$GF37$$\"3[++]i`1h\"*F0$\"3]**\\7oK0eGF37$$\"3W++]P
?Wl&*F0$\"3%***\\(=5s#yGF37$$\"\"\"F*$\"#HF*-%'COLOURG6&%$RGBG$\"#5F)$
F*F*Fa[l-%+AXESLABELSG6$Q\"h6\"Q!Ff[l-%&TITLEG6#%enGraph~of~(V(2~+~h)~
-~V(2))/h~over~the~interval~h~=~-1~to~1G-%%VIEWG6$;F(Fgz%(DEFAULTG" 1 
2 0 1 10 0 2 9 1 4 2 1.000000 46.000000 45.000000 0 0 "Curve 1" }}}}
{SECT 0 {PARA 4 "" 0 "" {TEXT -1 0 "" }{TEXT 275 30 " b) Evaluate the \+
average rates" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "evalf((V(1
.99) - V(2))/(1.99 - 2));\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"%&R#
!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "evalf((V(2.01) - V(
2))/(2.01 - 2));   \n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"%0C!\"#" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "evalf((V(2.01) - V(1.99))/
(2.01 - 1.99));  \n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\")+++C!\"'" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 4 "" 0 "
" {TEXT -1 1 " " }{TEXT 276 59 "c)  Evaluate the instantaneous rate us
ing appropriate limit" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "limit((V(2 + h) - V(2))/h,
 h = 0);\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#C" }}}{SECT 0 {PARA 
4 "" 0 "" {TEXT -1 0 "" }{TEXT 277 45 " d) Evaluate the rate using a t
able of values" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "L := [h, \+
`   `,      evalf((V(2 + h) - V(2))/h)];\n" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"LG7%%\"hG%$~~~G*&,(*&$\"\"&\"\"!\"\"\"),&$\"\"#F-F.
F&F.F2F.F.$\"#?F-!\"\"*&$\"\"%F-F.F&F.F.F.F&F5" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 51 "for h from  -0.05 by .01 to -.01 do print (L);
 od;\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$!\"&!\"#%$~~~G$\"++++vB!
\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$!\"%!\"#%$~~~G$\"++++!Q#!\")
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$!\"$!\"#%$~~~G$\"++++&Q#!\")" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$!\"#F%%$~~~G$\"++++!R#!\")" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$!\"\"!\"#%$~~~G$\"++++&R#!\")" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "for  h  from  0.01 by .01 to
 .05 do print (L); od;\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$\"\"\"!
\"#%$~~~G$\"++++0C!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$\"\"#!\"#
%$~~~G$\"++++5C!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$\"\"$!\"#%$~
~~G$\"++++:C!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$\"\"%!\"#%$~~~G
$\"++++?C!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7%$\"\"&!\"#%$~~~G$\"
++++DC!\")" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "
" {TEXT -1 0 "" }{TEXT 278 15 " e) Conclusion " }}}{PARA 0 "" 0 "" 
{TEXT -1 85 "     The estimates for the instantaneous rate at t = 2 in
  a),  b),  c), and  d) are " }}{PARA 0 "" 0 "" {TEXT -1 28 "the same \+
and equal to 24.   " }}}{PARA 0 "" 0 "" {TEXT -1 58 "_________________
_________________________________________" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 258 "" 0 "" {TEXT 272 1 "A" }{TEXT 269 9 "SSIGNMENT" }
{TEXT 297 1 " " }}{PARA 260 "" 0 "" {TEXT -1 37 "                     \+
                " }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 1 " " }{TEXT 271 
9 "Problem: " }}{PARA 0 "" 0 "" {TEXT -1 81 "A stone is dropped into a
 pool of water and causes a ripple that travels in the  " }}{PARA 0 "
" 0 "" {TEXT -1 69 "shape of a circle out from the point of impact at \+
a rate of 2 m/s.   " }}{PARA 0 "" 0 "" {TEXT -1 83 "a)  Find the formu
la for the average change in the area, A(t), within the circle.  " }}
{PARA 0 "" 0 "" {TEXT -1 67 "     Define the average rate of change in
 area from t = 3 to 3 + h." }}{PARA 0 "" 0 "" {TEXT -1 81 "b)  Plot th
e graph of the average rate of change in the area from t = 3 to 3 + h \+
" }}{PARA 0 "" 0 "" {TEXT -1 69 "     over h = -1 to 1.  Use it to est
imate the rate increase at t =3." }}{PARA 0 "" 0 "" {TEXT -1 59 "c)  W
hat is the average rate of increase in the area from  " }}{PARA 0 "" 
0 "" {TEXT -1 83 "      t = 2.99 to t = 3.00,  t = 3.00 to t = 3.01,  \+
t = 2.99 to t = 3.01 minutes.  " }}{PARA 0 "" 0 "" {TEXT -1 64 "      \+
        Use this to estimate the rate of incease at t = 3." }}{PARA 0 
"" 0 "" {TEXT -1 81 "d)  Find the instantaneous rate of increase at t \+
= 3 by taking appropriate limit " }}{PARA 0 "" 0 "" {TEXT -1 18 "     \+
using Maple. " }}{PARA 0 "" 0 "" {TEXT -1 72 "e)  Find the instantaneo
us rate of increase at t = 3 by making a table  " }}{PARA 0 "" 0 "" 
{TEXT -1 81 "             for  (A(3 + h) - A(3))/h for values of h clo
se to 0 omitting h = 0. " }}{PARA 0 "" 0 "" {TEXT -1 2 "f)" }{TEXT 
257 1 " " }{TEXT -1 45 " Compare your answer with the answers in  b),
" }{TEXT 258 2 "  " }{TEXT -1 3 "c)," }{TEXT 259 1 " " }{TEXT -1 2 "d)
" }{TEXT 260 1 "," }{TEXT -1 5 " and " }{TEXT 279 1 " " }{TEXT -1 3 "e
)." }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 58 " _
________________________________________________________" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 104 "MSIP Grant #P120A
80089-98:  \"Three Urban Calculus Reform programs: Adopting the Best\"
 1998-2001         " }}}{MARK "13 0 1" 9 }{VIEWOPTS 1 1 0 1 1 1803 1 
1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }